Need Some Math Help

Training Advice/Discussion General

5 years ago 09/21/2018 12:39pm EDT

I used to be good at math back in the day but I haven't been in school for 10+ years so I am very rusty on my algebra. This equation is usually used for finding your estimated time when comparing races of different distances, however I would like to use it instead to find the distance you could run in a specific time.

The equation is:

t2 = t1 * (d2 / d1)^1.06

Where t1 is your current time for the given distance d1. And d2 is the new race distance, t2 is the time you are solving for.

How would I rearrange this equation to solve for d2 instead of solving for t2? I have forgotten how powers work when solving equations. Currently I have t2/t1 = (d2/d1)^1.06. I would like to manipulate the equation so I am left with only having d2 on one side of the equation.

TLDR

Given t2 = t1 * (d2/d1)^1.06 solve for d2

5 years ago 09/21/2018 12:42pm EDT

re: basic algebra

Why do you need to rearrange the equation? Just put in the data you have and leave d2 as x. Solve.

5 years ago 09/21/2018 12:45pm EDT

re: JustAnotherRunner2001

JustAnotherRunner2001 wrote:
Why do you need to rearrange the equation? Just put in the data you have and leave d2 as x. Solve.

This can't be a guess and check thing, I need to use this equation several times for several different athletes. It would take a long time to just plug in numbers until they work.

5 years ago 09/21/2018 12:50pm EDT

re: basic algebra

Raise both sides to the inverse of your current power. i.e. 1/1.06 = 0.943. Raise both sides to the 0.943 power. A power raised to a power is the same as multiplying the two powers together.

So the next step in your equation would be (t2/t1)^0.943 = d2/d1.

Then multiply both sides by d1:

d2 = d1 * (t2/t1)^0.943

If it helps to think of this with more recognizable powers, think of squaring something vs. taking the square root. Raising to the power of 1/2 is the same as taking square root, i.e. sqrt(x) = x^0.5. If I take the square root of x^2, it's the same as saying (x^2)^0.5, which is the same as x^(2 * 0.5) = x^1 = x.

5 years ago 09/21/2018 12:56pm EDT

re: Mikey A

Mikey A wrote:
Raise both sides to the inverse of your current power. i.e. 1/1.06 = 0.943. Raise both sides to the 0.943 power. A power raised to a power is the same as multiplying the two powers together.
So the next step in your equation would be (t2/t1)^0.943 = d2/d1.
Then multiply both sides by d1:
d2 = d1 * (t2/t1)^0.943
If it helps to think of this with more recognizable powers, think of squaring something vs. taking the square root. Raising to the power of 1/2 is the same as taking square root, i.e. sqrt(x) = x^0.5. If I take the square root of x^2, it's the same as saying (x^2)^0.5, which is the same as x^(2 * 0.5) = x^1 = x.

Awesome, thanks

5 years ago 09/21/2018 1:15pm EDT

re: basic algebra

Please don't use this shitty formula for comparing anyone's (especailly high schoolers') times...

5 years ago 09/21/2018 1:37pm EDT

re: basic algebra

Why on earth would you want your equation in this form? You'll need to know d2 to have a value for t2.

The only time you'd know t2 but not d2 would be if you were trying to figure out how far you could run in x minutes.

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